find the equation of contact plane with the chart: z = ln(1+xy) in the point (1,2,ln3). You also have to find the equation and parametrization normal line on the chart in given point. Solution F(x,y,z) = z - ln(1+xy) dF/dx = -y/(1+xy) = -2/(1+2) = -2/3 dF/dy = -x/(1+xy) = -1/(1+2) = -1/3 dF/dz = 1 Therefore the contact plane equation is: -2/3 (x - 1) - 1/3 (y - 2) + 1(z - ln(3)) = 0 or: 2x + y - 3z = 2 + 2 - 3ln(3) 2x + y - 3z = 4 - ln(3).